如何在JavaScript函数中传递一个php代码？

Question

我想将上传的文件名添加到（'.list'）中。文件的名称必须是它在服务器上传时被调用的名称。例如，我可以有2个文件，但有一个被称为mountains.png，另一个是mountains2.png。

但问题是，我怎么能传递$ _FILES [“fileImage”] [“name”]作为我的js函数的参数，然后追加它，因为JavaScript函数和PHP脚本是在单独的页面上PHP脚本确实做了一个回调函数）？

UPDATE

下面是JavaScript代码：

下面是形式代码（QandATable.php）

<form action='imageupload.php' method='post' enctype='multipart/form-data' target='upload_target' onsubmit='startImageUpload(this);' class='imageuploadform' > 
     <p>Image File: <input name='fileImage' type='file' class='fileImage' /> 
     <input type='submit' name='submitImageBtn' class='sbtnimage' value='Upload' /> 
     </p> 
     <ul class='list'></ul> 
     </form> 

下面是JavaScript函数（QandATable.php）

 function stopImageUpload(success){ 

    var nameimagefile = <?php echo $nameimagefile?>; 
     var result = ''; 
     if (success == 1){ 
     result = '<span class="msg">The file was uploaded successfully!</span><br/><br/>'; 
     $('.listImage').append(nameimagefile + '<br/>'); 
     } 
     else { 
     result = '<span class="emsg">There was an error during file upload!</span><br/><br/>'; 
     } 

    return true; 

    } 

下面是t他的PHP脚本（imageupload.php）：

$result = 0; 
    $nameimagefile = ''; 

     if(file_exists("ImageFiles/".$_FILES['fileImage']['name'])) { 
      $parts = explode(".",$_FILES['fileImage']['name']); 
      $ext = array_pop($parts); 
      $base = implode(".",$parts); 
      $n = 2; 

      while(file_exists("ImageFiles/".$base."_".$n.".".$ext)) $n++; 
      $_FILES['fileImage']['name'] = $base."_".$n.".".$ext; 

      move_uploaded_file($_FILES["fileImage"]["tmp_name"], 
      "ImageFiles/" . $_FILES["fileImage"]["name"]); 
      $result = 1; 
$nameimagefile = $_FILES["fileImage"]["name"]; 

     } 
      else 
       { 
       move_uploaded_file($_FILES["fileImage"]["tmp_name"], 
       "ImageFiles/" . $_FILES["fileImage"]["name"]); 
       $result = 1; 
$nameimagefile = $_FILES["fileImage"]["name"]; 

       } 


     ?> 

     <script language="javascript" type="text/javascript">window.top.window.stopImageUpload(<?php echo $result;?>);</script> 

Answer 1

你可以简单地把价值$ _FILE文件名到一个PHP变量比使用

var yourjasvariable=<?php echo $yourvariable?>; 

呼应，并使用这个js在append方法变量。 :-)

Answer 2

你可以选择AJAX做你想做的。 用JSON编写数据。 JSON可以从PHP和JavaScript 阅读 - 阅读JSON在PHP 获取数据 - 读取AJAX结果（JSON）从PHP

获取数据

我会做这样的事情（未经测试的例子）

AJAX JS部分

<form method='post' enctype='multipart/form-data' onsubmit='startAjaxImageUpload(this);' > 
     ... 
</form> 

/* 
* ajax functions 
*/ 
function startAjaxImageUpload(event){ 

    /* Collect your formdatas as json with jquery this datas will be sent to php*/ 
    var formDatas = { 
      'value1'  : $('input[test1=eid]').val(), 
      'value2'  : $('input[id=test2_id]').val(), 
      ...... 
     'value3' : $('input[id=test3_id]').val() 
     }; 

    $.ajax({ 
     cache: false, 
     url: "imageupload", 
     data: formDatas, 
     success: function(data) { 
      // data is the json Result from php => imageupload.php do what u want with them in js 
      // use the next line if u wanna see which json datas comes back from php if the ajax call wass successfull 
      // console.log("data is %o, data); 
      // .... 

     } 
     error:function(data){ 
      // error function 
      // data is the json Result from php => imageupload.php do what u want with them in js 
      // use the next line if u wanna see which json datas comes back from php if the ajax call wass successfull 
      // console.log("data is %o, data); 
      alert(damn, something went wrong); 
     } 
    }) 
} 

PHP的一部分，imageupload.php

$result = 0; 
    $nameimagefile = ''; 
    ..... 
    // if done ure work on server side and no error was found, pass the result back to starAjaxImageUpload success function 
    return $nameimagefile = $_FILES["fileImage"]["name"]; 
    }else 
    // abbort ajax, ajax error function will used 
    return false 
      }